A conjecture for primes $p\equiv\pm1\pmod5$

Question

For any prime $p\equiv\pm1\pmod5$, we can write $p$ uniquely in the form $x_p^2+3x_py_p+y_p^2$ with $x_p,y_p\in\mathbb Z$ and $x_p>y_p>0$.

I have the following conjecture.

Conjecture. We have $$\lim_{N\to+\infty}\frac{\sum_{p\le N\atop p\equiv\pm1\pmod5}(3x_p^2+2x_py_p)} {\sum_{p\le N\atop p\equiv\pm1\pmod5}(3y_p^2+2x_py_p)}=4.\tag{$*$}$$

I have checked this conjecture via computation. For example, \begin{align}\sum_{p\le 2\times10^9\atop p\equiv\pm1\pmod5}(3x_p^2+2x_py_p)=&88916125007243531, \\\sum_{p\le 2\times10^9\atop p\equiv\pm1\pmod5}(3y_p^2+2x_py_p)=&22228898519387861, \\\frac{88916125007243531}{22228898519387861}\approx&4.000023885.\end{align}

Similarly, any prime $p\equiv1\pmod3$ can be written uniquely as $u_p^2+u_pv_p+v_p^2$ with $u_p,v_p\in\mathbb Z$ and $u_p>v_p>0$, and I observe that $$\lim_{N\to+\infty}\frac{\sum_{p\le N\atop p\equiv1\pmod3}(u_p^2+2u_pv_p)} {\sum_{p\le N\atop p\equiv1\pmod3}(v_p^2+2u_pv_p)}=2.$$ This is essentially provable in view of Hecke's equidistribution theorem.

QUESTION. How to prove the equality $(*)$?

Answer 1

Let us consider $\mathbb{Q}(\sqrt{5})$ as a subfield of $\mathbb{R}$. Let us also consider the positive fundamental unit $\epsilon:=(1+\sqrt{5})/2$, whose square generates the group of totally positive units. The conditions on $x_p$, $y_p$ can be rewritten as $$p=x_p^2+3x_py_p+y_p^2\qquad\Longleftrightarrow\qquad p=(x_p+\epsilon^2 y_p)(x_p+\epsilon^{-2} y_p),$$ $$x_p+\epsilon^{-2} y_p<x_p+\epsilon^2 y_p<\epsilon^2(x_p+\epsilon^{-2} y_p)\qquad\Longleftrightarrow\qquad p^{1/2}<x_p+\epsilon^2 y_p<p^{1/2}\epsilon.$$ By Hecke's theorem, $\log(x_p+\epsilon^2 y_p)$ is equidistributed in $[\log(p^{1/2}),\log(p^{1/2}\epsilon)]$, cf. Example 3 in Ch. XV, §5 of Lang: Algebraic number theory. From here we get by a similar argument as in my response here that $$\lim_{N\to+\infty}\frac{\sum_{p\le N\atop p\equiv\pm1\pmod5}(3x_p^2+2x_py_p)} {\sum_{p\le N\atop p\equiv\pm1\pmod5}(3y_p^2+2x_py_p)} =\frac{\int_0^1 (3x(t)^2+2x(t)y(t))\,dt}{\int_0^1 (3y(t)^2+2x(t)y(t))\,dt},$$ where the real functions $x,y:[0,1]\to\mathbb{R}$ are defined by the equations $$\epsilon^{-1}x(t)+\epsilon y(t)=\epsilon^{-t}\qquad\text{and}\qquad \epsilon x(t)+\epsilon^{-1}y(t)=\epsilon^t.$$ Solving for $x(t)$ and $y(t)$, the ratio of the two integrals turns out to be exactly $4$. This proves $(\ast)$.